Question: Evaluate the iterated integral. $ \int_{-\pi}^0 \left( \int_0^{2\pi} \cos(2x) - 2\sin(y) \, dx \right) dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $-2\pi$ (Choice C) C $4 \pi$ (Choice D) D $8\pi$
Evaluate the inner integral: $\begin{aligned} & \int_{-\pi}^0 \left( \int_0^{2\pi} \cos(2x) - 2\sin(y) \, dx \right) dy \\ \\ &= \int_{-\pi}^0 \left[ \dfrac{\sin(2x)}{2} - 2x\sin(y) \right]_0^{2\pi} dy \\ \\ &= \int_{-\pi}^0 -4\pi \sin(y) \, dy \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_{-\pi}^0 -4\pi \sin(y) \, dy &= 4\pi\cos(y) \bigg|_{-\pi}^0 \\ \\ &= 4\pi( \cos(0) - \cos(-\pi) ) \\ \\ &= 8\pi \end{aligned}$ The answer: $ \int_{-\pi}^0 \left( \int_0^{2\pi} \cos(2x) - 2\sin(y) \, dx \right) dy = 8\pi$